Consider the following functions  $f\left(x\right)=\tan \left(\pi x^3\right)\sqrt{9+{\tan }^2\pi x}\sqrt[3]{27+{\tan }^3\pi x}$  and $g\left(x\right)=\frac{{\cos }^{-1}x+{\tan }^{-1}x-\cos x\times {\cos }^{-1}x-\cos x\times {\tan }^{-1}x}{\left(1+x^2\right){\cot }^{-1}x}$
 . The digit at the tens place in the value of $\frac{1}{\pi }\left(f\text{'}\left(-2\right)+\pi g\text{'}\text{'}\left(0\right)\right)$ is equal to

$$\begin{gathered} f\left(x\right)=\left[\frac{\tan \pi x^3}{u}\right]\left[\frac{\sqrt{9+{\tan }^{2}x}+\sqrt[3]{27+{\tan }^{3}x}}{v}\right]\Rightarrow f\text{'}\left(x\right)=u\text{'}v+uv\text{'} \\ u=0atx=-2\Rightarrow f\text{'}\left(-2\right)={\sec }^2\left(\pi x^3\right)3\pi x^2\sqrt{9+{\tan }^2\pi x}\sqrt[3]{27+{\tan }^3\pi x}atx=-2 \\ \therefore f\text{'}\left(-2\right)=12\pi 3.3=108\pi \\ g\left(x\right)=u.v\Rightarrow g\text{'}\text{'}\left(x\right)=u\text{'}\text{'}v+2u\text{'}v\text{'}+uv\text{'}\text{'};u=1-\cos x \\ atx=0;u=0,u\text{'}=0\Rightarrow g\text{'}\text{'}\left(0\right)=\cos x.{\left.\frac{{\cos }^{-1}x+{\tan }^{-1}x}{\left(1+x_{ }^2\right){\cot }^{-1}x}\right|}_{x=0}=1\times \frac{\frac{\pi }{2}+0}{1\times \frac{\pi }{2}}=1 \end{gathered}$$